### Understanding Probability: Birthday Problem#### Problem Statement:Assume that the year has 366 days, and all birthdays are equally likely. Rearrange the following steps in the correct order to find the probability that in a group of $ n $ people chosen at random, there are at least two born on the same day of the week.#### Explanation:This classic probability puzzle, commonly referred to as the "birthday problem," explores the likelihood of shared birthdays within a given group. By considering a year with 366 days, the calculations account for leap years, maintaining equal likelihood for each possible birthdate. This variant specifically examines the probability of shared birthdays occurring on the same day of the week within a randomly chosen group of $ n $ individuals.**Steps to Approach:**To solve, first consider the complementary probability—that no two individuals share the same birthday—and then use this to determine the desired probability. Rearrange the logical steps to build your solution:1. **Step 1: Calculate the total number of possible birthday assignments without restriction.**2. **Step 2: Determine the number of ways to assign unique birthdays.**3. **Step 3: Compute the complementary probability that no two individuals share the same birthday.**4. **Step 4: Subtract this probability from 1 to find the likelihood of at least one shared birthday.**By organizing these steps methodically, one grasps the underlying reasoning, equipping students to tackle similar probability problems effectively. ### Probability and BirthdaysUnderstanding the probability of events is a key concept in statistics. One interesting application of probability is the likelihood that at least two people in a group are born on the same day of the week. This probability involves sequential reasoning and combinatorial calculations.1. **Probability of Shared Birthdays:** The probability that at least two people in a group share the same birthday is given by: $$ \text{Probability that at least two are born on the same day of the week is therefore } 1 - p_n. $$ 2. **Probability of Distinct Birthdays:** To compute this, we first need the probability that all individuals in the group have different birthdays. This can be written as: $$ p_n = \frac{6}{7} \cdot \frac{5}{7} \cdot \ldots \cdot \frac{8-n}{7}. $$ Each term in this product represents a person in the group choosing a unique day of the week different from those chosen by the previous individuals.3. **Sequential Assignment:** Each person after the first must have a different birth day of the week from all the previous people in the group. Hence, the numerator 6 for the second person (since 1 day is taken), 5 for the third person (since 2 days are taken), and so forth up to $8-n$ for the nth person where they need to avoid the days already taken by the first $n-1$ people.Understanding these probabilities helps highlight the unexpected outcomes in probability theory and can be a compelling entry for students learning about statistics and permutations.

Answered: Image /qna-images/answer/15673fb6-1038-4f75-8616-113e985fb849.jpg

Assume that the year has 366 days and all birthdays are equally likely. Rearrange the following steps in the correct order to find the probability that in a group of n people chosen at andom, there are at least two born on the same day of the week.

Assume that the year has 366 days and all birthdays are equally likely. Rearrange the following steps in the correct order to find the probability that in a group of n people chosen at andom, there are at least two born on the same day of the week.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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